To calculate the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to consider the employees who eat breakfast, lunch, or both.
Total number of employees who eat breakfast = 20
Total number of employees who eat lunch = 50
Total number of employees who eat both breakfast and lunch = 10
Total number of employees who eat breakfast or lunch = 20 + 50 - 10 = 60
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is:
P(eats breakfast or lunch) = number of employees who eat breakfast or lunch / total number of employees surveyed
P(eats breakfast or lunch) = 60 / 80 = 0.75
So, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 0.75 or 75%.
A survey of 80 employees was conducted asking about which meals they regularly eat at the office. 20 employees eat breakfast at the office, 50 eat lunch, and 10 eat both breakfast and lunch. If an employee is randomly selected from this group, what is the probability of randomly selecting an employee who eat breakfast or lunch at the office?(2 points)
Responses
3 answers
a b c or d
The answer is b) 0.75 or 75%.